The point dividing a line in a given ratio

Edexcel International GCSE Further Pure Mathematics· 8B· 10 min read

1. The Section Formula for Internal Division★★☆☆☆⏱ 3 min

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When a point divides the line segment joining points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m : n$ , we use the section formula to find its coordinates. This formula is not provided in your exam, so you must memorize it.

📘 Definition

Section Formula

P\left(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}\right)

Gives coordinates of point P dividing segment AB from A to B in ratio $m : n$ , where $A P : P B = m : n$

Example:

For A(1, 2) and B(4, 6), ratio 1:2 gives $P (\frac{2 * 1 + 1 * 4}{3}, \frac{2 * 2 + 1 * 6}{3}) = (2, \frac{10}{3})$

📐 Worked Example

Find the coordinates of the point P that divides the line segment joining A(3, 5) and B(9, 2) in the ratio 2:1, where $A P : P B = 2 : 1$ .

1
Identify values for the formula: $x_{1} = 3, y_{1} = 5$ (coordinates of A), $x_{2} = 9, y_{2} = 2$ (coordinates of B), $m = 2, n = 1$
2
$x_P = \frac{n x_1 + m x_2}{m + n} = \frac{1*3 + 2*9}{2+1} = \frac{3 + 18}{3} = 7$
3
$y_P = \frac{n y_1 + m y_2}{m + n} = \frac{1*5 + 2*2}{3} = \frac{5 + 4}{3} = 3$
4
The coordinates of P are (7, 3)

2. Special Case: Midpoint of a Line Segment★☆☆☆☆⏱ 2 min

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The midpoint of a line segment is the point that divides the segment in the ratio 1:1, so we can substitute $m = 1$ and $n = 1$ into the section formula to get a simplified midpoint formula.

📘 Definition

Midpoint Formula

M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Gives the coordinates of the midpoint of the line segment joining points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$

Example:

Midpoint of (2, 4) and (6, 8) is $(\frac{2 + 6}{2}, \frac{4 + 8}{2}) = (4, 6)$

📐 Worked Example

Find the midpoint of the line segment joining C(-2, 7) and D(4, -3).

1
Identify values: $x_{1} = - 2, y_{1} = 7$ , $x_{2} = 4, y_{2} = - 3$
2
$x_M = \frac{-2 + 4}{2} = 1$
3
$y_M = \frac{7 + (-3)}{2} = 2$
4
Midpoint M has coordinates (1, 2)

✓ Quick check

What ratio corresponds to the midpoint of a line segment?
- 1:2
- 2:1
- 1:1
- 3:1
Reveal answer
2 —
Correct! The midpoint splits the segment into two equal parts, so the ratio of the two resulting segments is 1:1.

3. Solving for Unknown Endpoints or Ratios★★★☆☆⏱ 4 min

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You may be given the coordinates of the division point and one endpoint, and asked to find the other endpoint, or given both endpoints and the division point to find the ratio. Rearrange the section formula to solve for these unknown values.

📐 Worked Example

The point P(5, 3) divides the segment joining A(2, 9) and B(x, y) in the ratio 1:2 ( $A P : P B = 1 : 2$ ). Find the coordinates of B.

1
Substitute known values into the x-coordinate section formula: $m = 1, n = 2$ , $x_{1} = 2$ , $x_{P} = 5$
2
$5 = \frac{2*2 + 1*x}{1+2} = \frac{4 + x}{3}$
3
Multiply both sides by 3: $15 = 4 + x \to x = 11$
4
$Repeat for y-coordinate: $3 = \frac{2*9 + 1*y}{3} = \frac{18 + y}{3}$$
5
Multiply both sides by 3: $9 = 18 + y \to y = - 9$
6
Coordinates of B are (11, -9)

4. Common Pitfalls

Wrong move:

Swapping m and n when applying the section formula, e.g. multiplying m by the first point's coordinates instead of the second.

Why:

The ratio $m : n$ refers to $A P : P B$ , so m corresponds to the segment closer to the second endpoint B, not A, leading to inverted weights if swapped.

Correct move:

Label the ratio with the endpoints first: if $A P : P B = m : n$ , n multiplies A's coordinates, m multiplies B's coordinates before dividing by $m + n$ .

Wrong move:

Forgetting to divide by $m + n$ when calculating coordinates.

Why:

The section formula is a weighted average of the two endpoints, so omitting the division by the total ratio parts gives an incorrect scaled coordinate.

Correct move:

After calculating the weighted sum of x and y coordinates, always divide by $m + n$ , even for the midpoint formula (divide by $2 = 1 + 1$ ).

Wrong move:

Using the wrong order of endpoints when the ratio is given for B to A instead of A to B.

Why:

If the ratio is $P B : A P = m : n$ instead of $A P : P B$ , the values of m and n are swapped, leading to completely incorrect coordinates.

Correct move:

Rewrite the ratio explicitly in the order of the first named endpoint to the second named endpoint before substituting into the formula.

Wrong move:

Assuming the formula works for points outside the line segment AB.

Why:

All ratio division questions in 4PM1 only test internal division, with the point lying between the two endpoints; external division is not in the syllabus.

Correct move:

Confirm the division point is between A and B before applying the standard section formula, as external division will not be tested.

5. Quick Reference Cheatsheet

Scenario	Formula	Key Notes
Point dividing AB in ratio $m : n$ ( $A P : P B = m : n$ )	$(\frac{n x _{1} + m x _{2}}{m + n}, \frac{n y _{1} + m y _{2}}{m + n})$	A = $(x_{1}, y_{1})$ , B = $(x_{2}, y_{2})$
Midpoint of AB	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$	Special case $m : n = 1 : 1$
Solve for unknown endpoint/ratio	Rearrange formula to isolate unknown	Write one separate equation for x and y coordinates

6. Frequently Asked

Is the section formula provided on the 4PM1 formula sheet?

No, all coordinate geometry formulae including the section formula must be memorized for the exam, as they are not given in the formula booklet.

How do I remember which coordinates pair with m vs n?

If the ratio is written as AP:PB = m:n for segment AB, n multiplies coordinates of A (first endpoint) and m multiplies coordinates of B (second endpoint) before dividing by m + n.

Going deeper

official_documentEdexcel IGCSE Further Pure Math 4PM1 SpecificationRefer to section 8 for full coordinate geometry requirements

What's Next

Now that you can find points dividing lines in a given ratio, you are ready to move on to other core coordinate geometry topics in Edexcel IGCSE Further Pure Math. The next topic covers calculating the gradient of a line segment, which you will use alongside ratio division to solve full straight-line geometry problems. You can also practice applying this formula to structured exam questions to build speed and accuracy before your assessment. Make sure you memorize the section formula fully, as it is not provided in the exam formula booklet and will be required for nearly all coordinate geometry questions in section 8 of the syllabus.